Abraham Lincoln University Chemistry Spectroscopy Worksheet

CHEM0841 Spectroscopy worksheet: Interpreting Infrared and NMR spectraSection 1:
You are given four pairs of different spectra below. Making use of both Infrared (IR) and 1H Nuclear
Magnetic Resonance (NMR) spectroscopic data, match each compound to the correct spectra (A, B ,
C or D). In the case of each IR spectrum, identify diagnostic bands (in cm-1) that can be used to support
your assignment (for sharp bands estimate the exact position, for broad bands, estimate a range). In
the case of each NMR spectrum, label each peak as far as possible (e.g. CH3, NH2, aromatic-H, etc. (if
applicable)).
1
Compound A
2
Compound B
3
Compound C (Hint: 1H NMR spectrum – D2O shake below (Note: Adding D2O to the NMR tube results
in disappearance of OH, NH and NH2 signals)).
4
Compound D
5
Spectroscopy Workshop 26 March 2020
1. For each of the following five compounds A-E assign, giving your reasoning, each compound
to the correct spectrum 1-5. You should quote the position of all peaks (in cm-1) that support
your assignments (for sharp peaks estimate the exact position, for broad peaks, estimate a
range).
Spectrum 1
Spectrum 2
Spectrum 3
Spectrum 4
Spectrum 5
2.
Explain briefly the following observations from proton NMR spectra:
(a) chloromethane has a higher chemical shift than tert-butyl chloride;
(b) ethene and benzene have much higher chemical shifts than cyclohexane.
3.
For each of the compounds (i) to (iv):
(a) state how many different proton environments there are in the molecule, and
indicate these clearly on the structure;
(b) give an approximate value for the expected chemical shift for the signal
corresponding to each environment;
(c) give the expected integral for each signal.
4.
The 1H NMR spectra (A-C) of the three isomeric compounds (i)-(iii) shown below are
given underneath. Using your knowledge of chemical shift ranges and integrals, match
the spectra to the structures. Explain your reasoning.
A
9.5
9.0
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
B
2.5
9.5
9.0
8.5
8.0
7.5
7.0
6.5
2.0
6.0
5.5
5.0
1.5
4.5
4.0
1.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
C
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
CHEM0841 Spectroscopy workshop 26th March 2020 Answers
Answer 1
A5; B4; C2; D3; E1
Spectrum 1 – NH2 asymm/sym stretches, C=O amide stretch at around 1680 and NO2 stretches
Spectrum 2 – CN stretch and C=O acid chloride stretch
Spectrum 3 – CN stretch, V-shaped OH stretch and acid C=O
Spectrum 4 – CN stretch, C=O ester stretch
Spectrum 5 – V-shaped OH stretch, acid C=O and NO2 stretches
For C=O peaks, type and exact frequencies should be noted.
Answer 2
(a) Chloromethane comes around 3 ppm, t-butyl chloride around 1.5 ppm. Deshielded by
electronegative element.
(b) Deshielded by anisotropic effect due to -cloud. Magnetic field generated by -cloud is larger for
benzene hence additional shift
Answer 3
Answer 4
Spectra A & C have aldehyde peaks above 9 ppm. The integrals of the aromatic protons around 7-8
are A = 4, B = 4, C = 5. Spectrum C has two different peaks in the alkene region 5.0-6.0. A & B both
have a 2H and a 3H peak in the alkane region, whereas C has only a 3H peak.
This is sufficient to assign C = (i) (5 aromatic H and alkene). A & B still look quite similar, but the
chemical shifts of the methyl groups are distinctive. A is (iii) because the deshielded methyl ketone
comes around 2.1 ppm, not 1.3 ppm in B, which must therefore be (ii).